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Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
9,\\
\mathop {\lim }\limits_{x \to 4} \left( {\frac{{\sqrt {7x + 8} .\sqrt[3]{{9x - 28}} - 12}}{{{x^3} - 4{x^2} + 2x - 8}}} \right)\\
= \mathop {\lim }\limits_{x \to 4} \frac{{\sqrt {7x + 8} .\left( {\sqrt[3]{{9x - 28}} - 2} \right) + 2.\left( {\sqrt {7x + 8} - 6} \right)}}{{{x^3} - 4{x^2} + 2x - 8}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{\sqrt {7x + 8} .\frac{{{{\sqrt[3]{{9x - 28}}}^3} - {2^3}}}{{{{\sqrt[3]{{9x - 28}}}^2} + \sqrt[3]{{9x - 28}}.2 + {2^2}}} + 2.\frac{{{{\sqrt {7x + 8} }^2} - {6^2}}}{{\sqrt {7x + 8} + 6}}}}{{{x^2}\left( {x - 4} \right) + 2\left( {x - 4} \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{\sqrt {7x + 8} .\frac{{\left( {9x - 28} \right) - 8}}{{{{\sqrt[3]{{9x - 28}}}^2} + 2\sqrt[3]{{9x - 28}} + 4}} + 2.\frac{{\left( {7x + 8} \right) - 36}}{{\sqrt {7x + 8} + 6}}}}{{\left( {x - 4} \right)\left( {{x^2} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{\sqrt {7x + 8} .\frac{{9\left( {x - 4} \right)}}{{{{\sqrt[3]{{9x - 28}}}^2} + 2\sqrt[3]{{9x - 28}} + 4}} + 2.\frac{{7\left( {x - 4} \right)}}{{\sqrt {7x + 8} + 6}}}}{{\left( {x - 4} \right)\left( {{x^2} + 2} \right)}}\\
= \mathop {\lim }\limits_{x \to 4} \dfrac{{\frac{{9\sqrt {7x + 8} }}{{{{\sqrt[3]{{9x - 28}}}^2} + 2\sqrt[3]{{9x - 28}} + 4}} + \frac{{14}}{{\sqrt {7x + 8} + 6}}}}{{{x^2} + 2}}\\
= \dfrac{{\frac{{9\sqrt {7.4 + 8} }}{{{{\sqrt[3]{{9.4 - 28}}}^2} + 2\sqrt[3]{{9.4 - 28}} + 4}} + \frac{{14}}{{\sqrt {7.4 + 8} + 6}}}}{{{4^2} + 2}}\\
= \frac{{17}}{{54}}\\
10,\\
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {4x - 3} + \sqrt {2x - 1} - 3x + 1}}{{{x^2} - 2x + 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {4x - 3} - \left( {2x - 1} \right)} \right) + \left( {\sqrt {2x - 1} - x} \right)}}{{{x^2} - 2x + 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{{{\sqrt {4x - 3} }^2} - {{\left( {2x - 1} \right)}^2}}}{{\sqrt {4x - 3} + \left( {2x - 1} \right)}} + \dfrac{{{{\sqrt {2x - 1} }^2} - {x^2}}}{{\sqrt {2x - 1} + x}}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{4x - 3 - 4{x^2} + 4x - 1}}{{\sqrt {4x - 3} + 2x - 1}} + \frac{{2x - 1 - {x^2}}}{{\sqrt {2x - 1} + x}}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\frac{{ - 4{{\left( {x - 1} \right)}^2}}}{{\sqrt {4x - 3} + 2x - 1}} + \frac{{ - {{\left( {x - 1} \right)}^2}}}{{\sqrt {2x - 1} + x}}}}{{{{\left( {x - 1} \right)}^2}}}\\
= \mathop {\lim }\limits_{x \to 1} \left[ {\frac{{ - 4}}{{\sqrt {4x - 3} + 2x - 1}} + \frac{{ - 1}}{{\sqrt {2x - 1} + x}}} \right]\\
= \frac{{ - 4}}{{\sqrt {4.1 - 3} + 2.1 - 1}} + \frac{{ - 1}}{{\sqrt {2.1 - 1} + 1}}\\
= \frac{{ - 5}}{2}
\end{array}\)
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Hoa20492001ly giúp em bài 2.1 đc ko ạ em cần gấp rồi ạ